The Standard Atmosphere

Based on Chapter 3 of Introduction to Flight by John D. Anderson, Jr. and Mary L. Bowden (McGraw-Hill, 2021).

Aircraft performance depends critically on the properties of the air through which they fly. The standard atmosphere defines these properties — pressure, temperature, density, and viscosity — as functions of altitude, providing a common reference for all aerospace calculations.

1. Definition of Altitude

Anderson begins with a careful distinction between two types of altitude.

1.1 Geometric Altitude $h_G$

The geometric altitude is the actual height above sea level, measured along a straight line. It is the distance you would measure with a tape measure.

1.2 Geopotential Altitude $h$

The geopotential altitude accounts for the variation of gravity with height. Newton’s law of gravitation gives:

g = g_0 \left(\frac{r_e}{r_e + h_G}\right)^2

where $g_0 = 9.80665$ m/s² and $r_e = 6,356,766$ m (Earth’s effective radius at 45° latitude). The geopotential altitude is defined such that the work done against gravity equals $g_0 h$ :

g_0 h \equiv \int_0^{h_G} g \, dh_G

Substituting $g(h_G)$ and integrating:

\begin{aligned} g_0 h &= g_0 r_e^2 \int_0^{h_G} \frac{dh_G}{(r_e + h_G)^2} \\[4pt] &= g_0 r_e^2 \left[\frac{1}{r_e} - \frac{1}{r_e + h_G}\right] \\[4pt] &= g_0 \frac{r_e h_G}{r_e + h_G} \end{aligned}

Therefore:

\boxed{h = \frac{r_e}{r_e + h_G} h_G}

At $h_G = 65,000$ ft (≈ 20 km), the difference is only about 0.3%. For most aircraft performance work below 80,000 ft, $h \approx h_G$ is a safe approximation.

2. The Hydrostatic Equation

Consider a small fluid element of cross-sectional area $dA$ and height $dh_G$ :

Downward force: weight $W = \rho g (dA \, dh_G)$
Upward pressure force: $p \, dA$
Downward pressure force: $(p + dp) \, dA$

For a stationary fluid in equilibrium:

p \, dA = (p + dp) \, dA + \rho g (dA \, dh_G)

Simplifying:

\boxed{dp = -\rho g \, dh_G}

This is the hydrostatic equation — pressure decreases with altitude because the weight of the air column above decreases. In terms of geopotential altitude:

\boxed{dp = -\rho g_0 \, dh}

3. The Perfect Gas Equation of State

Air behaves as a perfect gas for the pressure and temperature ranges of the standard atmosphere:

\boxed{p = \rho R T}

where $R = 287.05287$ J/(kg·K) is the specific gas constant for dry air. For air, the ratio of specific heats is $\gamma = c_p/c_v = 1.4$ .

4. Temperature Distribution — The Lapse Rate Model

The standard atmosphere divides the atmosphere into gradient layers (temperature varies linearly) and isothermal layers (temperature constant).

4.1 Gradient Layers

In a gradient layer, temperature varies linearly with geopotential altitude:

\boxed{T = T_b + a (h - h_b)}

where $a = dT/dh$ is the lapse rate. A negative lapse rate means temperature decreases with altitude.

4.2 ISA Layer Definitions

Layer	$h$ (km)	$T_b$ (K)	$a$ (K/km)	$p_b$ (Pa)
Troposphere	0 — 11	288.15	−6.5	1.01325 × 10⁵
Tropopause	11 — 20	216.65	0	2.2632 × 10⁴
Stratosphere 1	20 — 32	216.65	+1.0	5.4749 × 10³
Stratosphere 2	32 — 47	228.65	+2.8	8.6802 × 10²
Stratopause	47 — 51	270.65	0	1.1091 × 10²
Mesosphere 1	51 — 71	270.65	−2.8	6.6939 × 10¹
Mesosphere 2	71 — 84.85	214.65	−2.0	3.9564 × 10⁰
Lower Thermosphere	84.85 — 91	186.95	0	2.1410 × 10⁻¹
Upper Thermosphere	91 — 105	186.95	+1.5	6.96 × 10⁻²

5. Pressure Integration — Gradient Layers

Combining the hydrostatic equation with the perfect gas law eliminates density:

\frac{dp}{p} = -\frac{g_0}{R} \frac{dh}{T}

For a gradient layer where $T = T_b + a(h - h_b)$ , substitute and integrate:

\begin{aligned} \int_{p_b}^{p} \frac{dp}{p} &= -\frac{g_0}{R} \int_{h_b}^{h} \frac{dh}{T_b + a(h - h_b)} \\[4pt] \ln\frac{p}{p_b} &= -\frac{g_0}{aR} \ln\frac{T}{T_b} \end{aligned}

Therefore, for gradient layers ( $a \neq 0$ ):

\boxed{\frac{p}{p_b} = \left(\frac{T}{T_b}\right)^{-g_0/(aR)}}

Example: Troposphere (0–11 km)

$a = -0.0065$ K/m, so the exponent is:

-\frac{g_0}{aR} = -\frac{9.80665}{(-0.0065)(287.05287)} = 5.256

Thus:

\frac{p}{p_b} = \left(\frac{T}{288.15}\right)^{5.256}

At 11 km: $T = 288.15 - 0.0065 \times 11,000 = 216.65$ K

\frac{p}{p_b} = \left(\frac{216.65}{288.15}\right)^{5.256} = 0.2234

So $p = 0.2234 \times 101,325 = 22,632$ Pa — pressure at the tropopause.

6. Pressure Integration — Isothermal Layers

For isothermal layers ( $a = 0$ , $T = T_b = \text{constant}$ ):

\frac{dp}{p} = -\frac{g_0}{RT_b} dh

Integrating:

\boxed{\frac{p}{p_b} = \exp\left[-\frac{g_0}{RT_b}(h - h_b)\right]}

The Scale Height

The quantity $H = RT_b/g_0$ is the atmospheric scale height. In isothermal layers:

p = p_b \exp\left(-\frac{h - h_b}{H}\right)

At the tropopause ( $T = 216.65$ K): $H = 287.05 \times 216.65 / 9.80665 = 6,342$ m. Every 6.34 km, pressure drops by $1/e \approx 37\%$ of its value.

7. Density Distribution

Once pressure and temperature are known, density follows from the perfect gas law:

\boxed{\rho = \frac{p}{RT}}

At any altitude, relative density is:

\frac{\rho}{\rho_b} = \frac{p/p_b}{T/T_b}

For gradient layers:

\frac{\rho}{\rho_b} = \left(\frac{T}{T_b}\right)^{-[g_0/(aR) + 1]}

For the troposphere, the density exponent is $-[5.256 + 1] = -6.256$ :

At 11 km:

\frac{\rho}{\rho_{SL}} = \left(\frac{216.65}{288.15}\right)^{-6.256} = \left(\frac{216.65}{288.15}\right)^{6.256} = 0.297

Density at the tropopause is only 29.7% of sea-level density, compared to 22.3% for pressure. Density drops faster than pressure because of the simultaneous temperature decrease.

8. Speed of Sound

From thermodynamics, the speed of sound in a perfect gas:

\boxed{a = \sqrt{\gamma R T}}

For air ( $\gamma = 1.4$ , $R = 287.05$ J/(kg·K)):

a = \sqrt{1.4 \times 287.05 \times T} = 20.05 \sqrt{T} \text{ m/s}

Altitude	$T$ (K)	$a$ (m/s)	Mach 1 equivalent
Sea level	288.15	340.3	1,225 km/h
11 km	216.65	295.1	1,062 km/h
20 km	216.65	295.1	1,062 km/h
47 km	270.65	329.8	1,187 km/h
85 km	186.95	274.1	987 km/h

9. Viscosity — Sutherland’s Law

Anderson presents Sutherland’s law for the temperature dependence of viscosity:

\boxed{\frac{\mu}{\mu_0} = \left(\frac{T}{T_0}\right)^{3/2} \frac{T_0 + S}{T + S}}

where $S = 110.4$ K (Sutherland’s constant for air) and $\mu_0 = 1.7894 \times 10^{-5}$ Pa·s at $T_0 = 288.15$ K.

In the troposphere, as temperature drops, viscosity decreases — at 11 km, $\mu \approx 1.42 \times 10^{-5}$ Pa·s.

10. Pressure, Density, and Temperature Altitudes

Anderson defines three important derived altitudes:

Pressure Altitude

The altitude in the standard atmosphere corresponding to a given pressure. If a pressure gauge reads 70 kPa, the pressure altitude is the ISA altitude where $p = 70$ kPa. This is what an altimeter displays when set to 29.92 inHg.

Density Altitude

The altitude in the standard atmosphere corresponding to a given density:

h_d = h + \frac{T_0}{a}\left[1 - \left(\frac{\rho_{actual}}{\rho_0}\right)^{-1/[g_0/(aR)+1]}\right]

High density altitude (hot day, high elevation) degrades aircraft performance because the engine produces less power and the wings produce less lift.

Temperature Altitude

The altitude in the standard atmosphere where $T = T_{actual}$ . In the troposphere:

h_T = \frac{T_0 - T_{actual}}{|a|}

ISA Properties Calculator

Enter an altitude (0–105 km) and click Calculate to view the ISA temperature profile. Hover over the curve for exact values at any altitude.

Altitude (km):

● Gradient layer● Isothermal layer- - - Layer boundary◎ 0 °C (273.15 K)

This interactive plot requires a larger screen.
Please view this page on a computer to explore the ISA temperature profile.

Quick Reference — Key ISA Values

Altitude (km)	$T$ (K)	$p/p_{SL}$	$\rho/\rho_{SL}$	$a$ (m/s)	Layer
0	288.2	1.000	1.000	340.3	Troposphere
5	255.7	0.533	0.601	320.6	Troposphere
11	216.7	0.223	0.297	295.1	Tropopause
20	216.7	0.054	0.072	295.1	Stratosphere
32	228.7	0.0086	0.011	303.1	Stratosphere
47	270.7	0.0011	0.0012	329.8	Stratopause
51	270.7	0.00066	0.00070	329.8	Mesosphere
71	214.7	0.000039	0.000053	293.7	Mesosphere
85	187.0	0.0000037	0.0000058	274.1	Thermosphere
105	208.0	0.00000034	0.00000047	289.0	Thermosphere

Notes:

$p_{SL} = 101,325$ Pa, $\rho_{SL} = 1.225$ kg/m³
At 11 km: ~75% of the atmosphere’s mass lies below this altitude
At 47 km: pressure is ~0.1% of sea level — 99.9% of the atmosphere is below
The lapse rate exponent in the troposphere is 5.256 — memorise this number for exams
Density drops faster than pressure due to simultaneous temperature decrease
The coldest point in the atmosphere is the mesopause at ~85 km (−86 °C)
All data follows the ISO 2533:1975 / U.S. Standard Atmosphere 1976 model

Summary of Key Equations

Anderson’s Chapter 3 provides the mathematical foundation for all later chapters on aerodynamics and aircraft performance. The key results are:

Equation	Formula	Use
Hydrostatic	$dp = -\rho g_0 dh$	Foundation of atmospheric model
Geopotential altitude	$h = \frac{r_e}{r_e + h_G} h_G$	Relates geopotential to geometric altitude
Gradient layer pressure	$\frac{p}{p_b} = \left(\frac{T}{T_b}\right)^{-g_0/(aR)}$	Pressure in troposphere, stratosphere, mesosphere
Isothermal layer pressure	$\frac{p}{p_b} = \exp\left[-\frac{g_0}{RT_b}(h-h_b)\right]$	Pressure in tropopause, stratopause
Density	$\rho = p/(RT)$	Derived from pressure and temperature
Speed of sound	$a = \sqrt{\gamma R T}$	Mach number calculations
Sutherland’s law	$\frac{\mu}{\mu_0} = \left(\frac{T}{T_0}\right)^{3/2} \frac{T_0 + S}{T + S}$	Viscosity at any altitude

References

Anderson, J.D. and Bowden, M.L. — Introduction to Flight, 9th Edition, Chapter 3, McGraw-Hill, 2021
U.S. Standard Atmosphere, 1976 — NOAA/NASA/USAF
ISO 2533:1975 — Standard Atmosphere

All derivations follow Anderson’s notation and methodology. Geopotential altitude $h$ is used throughout unless geometric altitude $h_G$ is explicitly noted.